When a concentrated solution of sulphanilic acid and 1-naphthylamine is treated with nitrous acid (273 K) and acidified with acetic acid, the mass (g) of 0.1 mole of product formed is :
(Given molar mass in ${\mathrm{gmol}}^{-1}\mathrm{H}:1,\mathrm{C}:12,\mathrm{N}:14$, $O:16,S:32)$

[JEE Main 2025, 2 Apr (Shift 2)]

Question

When a concentrated solution of sulphanilic acid and 1-naphthylamine is treated with nitrous acid (273 K) and acidified with acetic acid, the mass (g) of 0.1 mole of product formed is :
(Given molar mass in ${\mathrm{gmol}}^{-1}\mathrm{H}:1,\mathrm{C}:12,\mathrm{N}:14$, $O:16,S:32)$

[JEE Main 2025, 2 Apr (Shift 2)]

YESPYQ · Accepted Answer

Correct answer: c) $33$. Sulphanilic acid forms diazonium salt with nitrous acid at 273 K, which couples with 1-naphthylamine to form an azo dye. Molar mass of product = 327 g ${\mathrm{mol}}^{-1}$ Mass of 0.1 mol = 33 g (approximately)

When a concentrated solution of sulphanilic acid and 1-naphthylamine is treated with nitrous acid (273 K) and acidified …

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